Related
I have written a simple C code for pointers. As per my understanding, Pointer is a variable which holds the address of another variable.
Eg :
int x = 25; // address - 1024
int *ptr = &x;
printf("%d", *ptr); // *ptr will give value at address of x i.e, 25 at 1024 address.
However when I try below code I'm getting segmentation fault
#include "stdio.h"
int main()
{
int *ptr = 25;
printf("%d", *ptr);
return 0;
}
What's wrong in this? Why can't a pointer variable return the value at address 25? Shouldn't I be able to read the bytes at that address?
Unless you're running on an embedded system with specific known memory locations, you can't assign an arbitrary value to a pointer an expect to be able to dereference it successfully.
Section 6.5.3.2p4 of the C standard states the following regarding the indirection operator *:
The unary
* operator denotes indirection. If the operand points to a function, the result is a function designator; if it points to an
object, the result is an lvalue designating the object. If
the operand has type "pointer to type", the result has
type "type". If an invalid value has been assigned to
the pointer, the behavior of the unary
* operator is undefined.
As mentioned in the passage above, the C standard only allows for pointers to point to known objects or to dynamically allocated memory (or NULL), not arbitrary memory locations. Some implementations may allow that in specific situations, but not in general.
Although the behavior of your program is undefined according to the C standard, your code is actually correct in the sense that it is doing exactly what you intend. It is attempting to read from memory address 25 and print the value at that address.
However, in most modern operating systems, such as Windows and Linux, programs use virtual memory and not physical memory. Therefore, you are most likely attempting to access a virtual memory address that is not mapped to a physical memory address. Accessing an unmapped memory location is illegal and causes a segmentation fault.
Since the memory address 0 (which is written in C as NULL) is normally reserved to specify an invalid memory address, most modern operating systems never map the first few kilobytes of virtual memory addresses to physical memory. That way, a segmentation fault will occur when an invalid NULL pointer is dereferenced (which is good, because it makes it easier to detect bugs).
For this reason, you can be reasonably certain that also the address 25 (which is very close to address 0) is never mapped to physical memory and will therefore cause a segmentation fault if you attempt to access that address.
However, most other addresses in your program's virtual memory address space will most likely have the same problem. Since the operating system tries to save physical memory if possible, it will not map more virtual memory address space to physical memory than necessary. Therefore, trying to guess valid memory addresses will fail, most of the time.
If you want to explore the virtual address space of your process to find memory addresses that you can read without a segmentation fault occuring, you can use the appropriate API supplied by your operating system. On Windows, you can use the function VirtualQuery. On Linux, you can read the pseudo-filesystem /proc/self/maps. The ISO C standard itself does not provide any way of determining the layout of your virtual memory address space, as this is operating system specific.
If you want to explore the virtual memory address layout of other running processes, then you can use the VirtualQueryEx function on Windows and read /proc/[pid]/maps on Linux. However, since other processes have a separate virtual memory address space, you can't access their memory directly, but must use the ReadProcessMemory and WriteProcessMemory functions on Windows and use /proc/[pid]/mem on Linux.
Disclaimer: Of course, I don't recommend messing around with the memory of other processes, unless you know exactly what you are doing.
However, as a programmer, you normally don't want to explore the virtual memory address space. Instead, you normally work with memory that has been assigned to your program by the operating system. If you want the operating system to give you some memory to play around with, which you are allowed to read from and write to at will (i.e. without segmentation faults), then you can just declare a large array of chars (bytes) as a global variable, for example char buffer[1024];. Be careful with declaring larger arrays as local variables, as this may cause a stack overflow. Alternatively, you can ask the operating system for dynamically allocated memory, for example using the malloc function.
You should consider all warnings that the compiler issues.
This statement
int *ptr = 25;
is incorrect. You are trying to assign an integer to a pointer as an address of memory. Thus in this statement
printf("%d", *ptr);
there is an attempt to access memory at address 25 that does not belong to your program.
What you mean is the following
#include "stdio.h"
int main( void )
{
int x = 25;
int *ptr = &x;
printf("%d", *ptr);
return 0;
}
Or
#include "stdio.h"
#include <stdlib.h>
int main( void )
{
int *ptr = malloc( sizeof( int ) );
*ptr = 25;
printf("%d", *ptr);
free( ptr );
return 0;
}
why below program taking memory difference to next location fix 12 in next memory location or its have some logic for char/float /int have diffrencet different -2 difference to next location
include
#include
int main () {
char var;
char *ptr;
char **pptr;
char ***ppptr;
char ****pppptr;
var = 'M';
/* take the address of var */
ptr = &var;
/* take the address of ptr using address of operator & */
pptr = &ptr;
ppptr=&pptr;
ppptr=&pptr;
ppptr=&pptr;
pppptr=&ppptr;
/* take the value using pptr */
printf("Value of var = %d\n", var );
printf("Value available at *ptr = %c....and address is %u\n", *ptr ,ptr);
printf("Value available at **pptr = %c....and address is %u\n", **pptr,pptr);
printf("Value available at ***ppptr = %c....and address is %u\n", ***ppptr ,ppptr);
printf("Value available at ****pppptr = %c....and address is %u\n", ****pppptr,pppptr);
getch();
return 0;
}
//out put
///-----------------------------------------------
Value of var = 77;
Value available at *ptr = M....and address is 11401207
Value available at **pptr = M....and address is 11401192 why differnceis 12 in memory locaton
Value available at ***ppptr = M....and address is 11401180
Value available at ****pppptr = M....and address is 11401168
You appear to be asking about why the address of ptr, as judged by printing the value of pptr using scanf field descriptor %u, appears to differ from the address of the next-declared local variable, pptr, by 12 bytes. And so on down the line for the other local variables, I presume.
In fact, as #SouravGhosh already described in his answer, the %u field descriptor does not correspond properly to pointer arguments, and therefore the effects of your scanf calls, including all aspects of the output, are undefined.
However, the actual results don't really matter for the purposes of the question. The C language provides no justification for assuming any particular ordering or difference between the addresses of any function's local variables. It is at the discretion of the compiler. There is therefore no general answer to the question.
If we hypothesize that the output you observed happens to accurately reflect the differences between the addresses of the local variables -- which is by no means certain -- then that would suggest that the compiler is intentionally leaving padding between the variables. It is entirely free to do so, for any purpose whatever. For example, the compiler might be trying to minimize the potential impact of off-by-one errors in code that accesses the local variables via the pointers to them.
First, a proper way to print pointers is with %p and with the pointer converted to void * when passed to printf, as in printf("%p\n", (void *) ptr);. Since you mention in a comment having used %P, differing from the %u shown in the question, I will presume that incorrect case is a typo, and you have reproduced the problem using %p with results similar to those shown in the question.
In that case, what is likely happening is that the compiler is storing the pointers ptr, pptr, ppptr, and pppptr at addresses that happen to be 12 bytes apart. Of course, it could store them merely four or eight bytes apart if pointers in your C implementation happened to be four or eight bytes in size. However, the compiler may be storing other things between the pointers, perhaps addresses of the strings you use in the printf statements or other incidental values. This makes me suspect that you compiled without optimization enabled. If you recompile with optimization, you will likely see different results.
In any case, a better explanation of the results would likely be had by asking the compiler to show the assembly code it generates (as by using the -S switch with GCC).
Using %u for printing pointer address results in undefined behavior.
Use %p format specifier and cast the argument to void *.
Is it possible to assign a variable the address you want, in the memory?
I tried to do so but I am getting an error as "Lvalue required as left operand of assignment".
int main() {
int i = 10;
&i = 7200;
printf("i=%d address=%u", i, &i);
}
What is wrong with my approach?
Is there any way in C in which we can assign an address we want, to a variable?
Not directly.
You can do this though : int* i = 7200;
.. and then use i (ie. *i = 10) but you will most likely get a crash. This is only meaningful when doing low level development - device drivers, etc... with known memory addreses.
Assuming you are on an x86-type processer on a modern operating system, it is not possible to write to aribtray memory locations; the CPU works in concert with the OS to protect memory so that one process cannot accidentally (or intentionally) overwrite another processes' memory. Allowing this would be a security risk (see: buffer overflow). If you try to anyway, you get the 'Segmentation fault' error as the OS/CPU prevents you from doing this.
For technical details on this, you want to start with 1, 2, and 3.
Instead, you ask the OS to give you a memory location you can write to, using malloc. In this case, the OS kernel (which is generally the only process that is allowed to write to arbitrary memory locations) finds a free area of memory and allocates it to your process. The allocation process also marks that area of memory as belonging to your process, so that you can read it and write it.
However, a different OS/processor architecture/configuration could allow you to write to an arbitrary location. In that case, this code would work:
#include <stdio.h>
void main() {
int *ptr;
ptr = (int*)7000;
*ptr = 10;
printf("Value: %i", *ptr);
}
C language provides you with no means for "attaching" a name to a specific memory address. I.e. you cannot tell the language that a specific variable name is supposed to refer to a lvalue located at a specific address. So, the answer to your question, as stated, is "no". End of story.
Moreover, formally speaking, there's no alternative portable way to work with specific numerical addresses in C. The language itself defines no features that would help you do that.
However, a specific implementation might provide you with means to access specific addresses. In a typical implementation, converting an integral value Ato a pointer type creates a pointer that points to address A. By dereferencing such pointer you can access that memory location.
Not portably. But some compilers (usually for the embedded world) have extensions to do it.
For example on IAR compiler (here for MSP430), you can do this:
static const char version[] # 0x1000 = "v1.0";
This will put object version at memory address 0x1000.
You can do in the windows system with mingw64 setup in visual studio code tool, here is my code
#include<stdio.h>
int main()
{
int *c;
c = (int *)0x000000000061fe14; // Allocating the address 8-bit with respect to your CPU arch.
*c = NULL; // Initializing the null pointer for allocated address
*c = 0x10; // Assign a hex value (you can assign integer also without '0x')
printf("%p\n",c); // Prints the address of the c pointer variable
printf("%x\n",*c); // Prints the assigned value 0x10 -hex
}
It is tested with mentioned environment. Hope this helps Happy coding !!!
No.
Even if you could, 7200 is not a pointer (memory address), it's an int, so that wouldn't work anyway.
There's probably no way to determine which address a variable will have. But as a last hope for you, there is something called "pointer", so you can modify a value on address 7200 (although this address will probably be inaccessible):
int *i = (int *)7200;
*i = 10;
Use ldscript/linker command file. This will however, assign at link time, not run time.
Linker command file syntax depends largely on specific compiler. So you will need to google for linker command file, for your compiler.
Approximate pseudo syntax would be somewhat like this:
In linker command file:
.section start=0x1000 lenth=0x100 myVariables
In C file:
#pragma section myVariables
int myVar=10;
It's not possible, maybe possible with compiler extensions. You could however access memory at an address you want (if the address is accessible to your process):
int addr = 7200;
*((int*)addr) = newVal;
I think '&' in &a evaluates the address of i at the compile time which i think is a virtual address .So it is not a Lvalue according to your compiler. Use pointer instead
following short c program:
void foo(int a, int b) {
printf("a = %p b = %p\n", &a, &b);
}
main() {
foo(1, 2);
}
ok, now I used gdb to view this program. I got as output:
a = 0x7fff5fbff9ac b = 0x7fff5fbff9a8
and stopped execution after the output (in foo()). now I examined 0x7fff5fbff9ac and the content was:
1....correct
then 0x7fff5fbff9a8 and the content:
2...correct
now I wanted to view the return address of the function and examined (a + 4 bytes) with:
x/g 0x7fff5fbff9b1 (8 bytes!! address, therefore "g" (giant word))
and its content was:
(gdb) x/g 0x7fff5fbff9b1
0x7fff5fbff9b1: 0xd700007fff5fbff9
BUT: THIS IS NOT THE RETURN ADR FROM MAIN! where is my fault?
There are a whole bunch of faulty assumptions in your question.
You're assuming that integer arguments are passed on the stack immediately above the return address (as they are in many--not all--x86 ABIs under the default calling conventions). If this were the case, then immediately following the call, your stack would look like this:
// stack frame of main( )
// ...
value of b
value of a
return address <--- stack pointer
However, your assumption is incorrect. You have compiled your code into a 64-bit executable (as evidenced by the size of the pointer you are printing). Per the OS X ABI, in a 64-bit Intel executable, the first few integer arguments are passed in register, not on the stack. Thus, immediately following the call, the stack actually looks like this:
// stack frame of main( )
// ...
return address <--- stack pointer
Since you take the address of a and b, they will be written to the stack at some point before the call to printf( ) (unless the compiler is really clever, and realizes that it doesn't actually need to hand printf( ) valid pointers because it won't use the value pointed to, but that would be pretty evil as optimizations go), but you really don't know where they will be relative to the return address; in fact, because the 64-bit ABI provides a red zone, you don't even know whether they're above or below the stack pointer. Thus, at the time that you print out the address of a and b, your stack looks like this:
// stack frame of main( )
// ...
return address |
// ... |
// a and b are somewhere down here | <-- stack pointer points somewhere in here
// ... |
In general, the C language standard says nothing about stack layouts, or even that there needs to be a stack at all. You cannot get this sort of information in any portable fashion from C code.
Firstly, &a + 4 is 0x7FFF5FBFF9B0, so you are looking one byte offset from where you think you are.
Secondly, the saved frame pointer lies between a and the return address, and this is the value you are seeing.
What you are doing wrong is making a whole bunch of incorrect and random assumptions about the layout of the stack frame on your given platform. Where did you get that weird idea about "a + 4 bytes" location supposedly holding the return address?
If you really want to do this, get the documentation for your platform (or do some reverse engineering) to find out where and how exactly the return address is stored. Making random guesses and then asking other people why your random guesses do not produce results you for some reason expect is not exactly a productive way to do it.
Just trying to understand how to address a single character in an array of strings. Also, this of course will allow me to understand pointers to pointers subscripting in general.
If I have char **a and I want to reach the 3rd character of the 2nd string, does this work: **((a+1)+2)? Seems like it should...
Almost, but not quite. The correct answer is:
*((*(a+1))+2)
because you need to first de-reference to one of the actual string pointers and then you to de-reference that selected string pointer down to the desired character. (Note that I added extra parenthesis for clarity in the order of operations there).
Alternatively, this expression:
a[1][2]
will also work!....and perhaps would be preferred because the intent of what you are trying to do is more self evident and the notation itself is more succinct. This form may not be immediately obvious to people new to the language, but understand that the reason the array notation works is because in C, an array indexing operation is really just shorthand for the equivalent pointer operation. ie: *(a+x) is same as a[x]. So, by extending that logic to the original question, there are two separate pointer de-referencing operations cascaded together whereby the expression a[x][y] is equivalent to the general form of *((*(a+x))+y).
You don't have to use pointers.
int main(int argc, char **argv){
printf("The third character of
argv[1] is [%c].\n", argv[1][2]);
}
Then:
$ ./main hello The third character of
argv[1] is [l].
That's a one and an l.
You could use pointers if you want...
*(argv[1] +2)
or even
*((*(a+1))+2)
As someone pointed out above.
This is because array names are pointers.
Theres a brilliant C programming explanation in the book Hacking the art of exploitation 2nd Edition by Jon Erickson which discusses pointers, strings, worth a mention for the programming explanation section alone https://leaksource.files.wordpress.com/2014/08/hacking-the-art-of-exploitation.pdf.
Although the question has already been answered, someone else wanting to know more may find the following highlights from Ericksons book useful to understand some of the structure behind the question.
Headers
Examples of header files available for variable manipulation you will probably use.
stdio.h - http://www.cplusplus.com/reference/cstdio/
stdlib.h - http://www.cplusplus.com/reference/cstdlib/
string.h - http://www.cplusplus.com/reference/cstring/
limits.h - http://www.cplusplus.com/reference/climits/
Functions
Examples of general purpose functions you will probably use.
malloc() - http://www.cplusplus.com/reference/cstdlib/malloc/
calloc() - http://www.cplusplus.com/reference/cstdlib/calloc/
strcpy() - http://www.cplusplus.com/reference/cstring/strcpy/
Memory
"A compiled program’s memory is divided into five segments: text, data, bss, heap, and stack. Each segment represents a special portion of memory that is set aside for a certain purpose. The text segment is also sometimes called the code segment. This is where the assembled machine language instructions of the program are located".
"The execution of instructions in this segment is nonlinear, thanks to the aforementioned high-level control structures and functions, which compile
into branch, jump, and call instructions in assembly language. As a program
executes, the EIP is set to the first instruction in the text segment. The
processor then follows an execution loop that does the following:"
"1. Reads the instruction that EIP is pointing to"
"2. Adds the byte length of the instruction to EIP"
"3. Executes the instruction that was read in step 1"
"4. Goes back to step 1"
"Sometimes the instruction will be a jump or a call instruction, which
changes the EIP to a different address of memory. The processor doesn’t
care about the change, because it’s expecting the execution to be nonlinear
anyway. If EIP is changed in step 3, the processor will just go back to step 1 and read the instruction found at the address of whatever EIP was changed to".
"Write permission is disabled in the text segment, as it is not used to store variables, only code. This prevents people from actually modifying the program code; any attempt to write to this segment of memory will cause the program to alert the user that something bad happened, and the program
will be killed. Another advantage of this segment being read-only is that it
can be shared among different copies of the program, allowing multiple
executions of the program at the same time without any problems. It should
also be noted that this memory segment has a fixed size, since nothing ever
changes in it".
"The data and bss segments are used to store global and static program
variables. The data segment is filled with the initialized global and static variables, while the bss segment is filled with their uninitialized counterparts. Although these segments are writable, they also have a fixed size. Remember that global variables persist, despite the functional context (like the variable j in the previous examples). Both global and static variables are able to persist because they are stored in their own memory segments".
"The heap segment is a segment of memory a programmer can directly
control. Blocks of memory in this segment can be allocated and used for
whatever the programmer might need. One notable point about the heap
segment is that it isn’t of fixed size, so it can grow larger or smaller as needed".
"All of the memory within the heap is managed by allocator and deallocator algorithms, which respectively reserve a region of memory in the heap for use and remove reservations to allow that portion of memory to be reused for later reservations. The heap will grow and shrink depending on how
much memory is reserved for use. This means a programmer using the heap
allocation functions can reserve and free memory on the fly. The growth of
the heap moves downward toward higher memory addresses".
"The stack segment also has variable size and is used as a temporary scratch pad to store local function variables and context during function calls. This is what GDB’s backtrace command looks at. When a program calls a function, that function will have its own set of passed variables, and the function’s code will be at a different memory location in the text (or code) segment. Since the context and the EIP must change when a function is called, the stack is used to remember all of the passed variables, the location the EIP should return to after the function is finished, and all the local variables used by that function. All of this information is stored together on the stack in what is collectively called a stack frame. The stack contains many stack frames".
"In general computer science terms, a stack is an abstract data structure that is used frequently. It has first-in, last-out (FILO) ordering
, which means the first item that is put into a stack is the last item to come out of it. Think of it as putting beads on a piece of string that has a knot on one end—you can’t get the first bead off until you have removed all the other beads. When an item is placed into a stack, it’s known as pushing, and when an item is removed from a stack, it’s called popping".
"As the name implies, the stack segment of memory is, in fact, a stack data structure, which contains stack frames. The ESP register is used to keep track of the address of the end of the stack, which is constantly changing as items are pushed into and popped off of it. Since this is very dynamic behavior, it makes sense that the stack is also not of a fixed size. Opposite to the dynamic growth of the heap, as the stack change
s in size, it grows upward in a visual listing of memory, toward lower memory addresses".
"The FILO nature of a stack might seem odd, but since the stack is used
to store context, it’s very useful. When a function is called, several things are pushed to the stack together in a stack frame. The EBP register—sometimes called the frame pointer (FP) or local base (LB) pointer
—is used to reference local function variables in the current stack frame. Each stack frame contains the parameters to the function, its local variables, and two pointers that are necessary to put things back the way they were: the saved frame pointer (SFP) and the return address. The
SFP is used to restore EBP to its previous value, and the return address
is used to restore EIP to the next instruction found after the function call. This restores the functional context of the previous stack
frame".
Strings
"In C, an array is simply a list of n elements of a specific data type. A 20-character array is simply 20 adjacent characters located in memory. Arrays are also referred to as buffers".
#include <stdio.h>
int main()
{
char str_a[20];
str_a[0] = 'H';
str_a[1] = 'e';
str_a[2] = 'l';
str_a[3] = 'l';
str_a[4] = 'o';
str_a[5] = ',';
str_a[6] = ' ';
str_a[7] = 'w';
str_a[8] = 'o';
str_a[9] = 'r';
str_a[10] = 'l';
str_a[11] = 'd';
str_a[12] = '!';
str_a[13] = '\n';
str_a[14] = 0;
printf(str_a);
}
"In the preceding program, a 20-element character array is defined as
str_a, and each element of the array is written to, one by one. Notice that the number begins at 0, as opposed to 1. Also notice that the last character is a 0".
"(This is also called a null byte.) The character array was defined, so 20 bytes are allocated for it, but only 12 of these bytes are actually used. The null byte Programming at the end is used as a delimiter character to tell any function that is dealing with the string to stop operations right there. The remaining extra bytes are just garbage and will be ignored. If a null byte is inserted in the fifth element of the character array, only the characters Hello would be printed by the printf() function".
"Since setting each character in a character array is painstaking and strings are used fairly often, a set of standard functions was created for string manipulation. For example, the strcpy() function will copy a string from a source to a destination, iterating through the source string and copying each byte to the destination (and stopping after it copies the null termination byte)".
"The order of the functions arguments is similar to Intel assembly syntax destination first and then source. The char_array.c program can be rewritten using strcpy() to accomplish the same thing using the string library. The next version of the char_array program shown below includes string.h since it uses a string function".
#include <stdio.h>
#include <string.h>
int main()
{
char str_a[20];
strcpy(str_a, "Hello, world!\n");
printf(str_a);
}
Find more information on C strings
http://www.cs.uic.edu/~jbell/CourseNotes/C_Programming/CharacterStrings.html
http://www.tutorialspoint.com/cprogramming/c_strings.htm
Pointers
"The EIP register is a pointer that “points” to the current instruction during a programs execution by containing its memory address. The idea of pointers is used in C, also. Since the physical memory cannot actually be moved, the information in it must be copied. It can be very computationally expensive to copy large chunks of memory to be used by different functions or in different places. This is also expensive from a memory standpoint, since space for the new destination copy must be saved or allocated before the source can be copied. Pointers are a solution to this problem. Instead of copying a large block of memory, it is much simpler to pass around the address of the beginning of that block of memory".
"Pointers in C can be defined and used like any other variable type. Since memory on the x86 architecture uses 32-bit addressing, pointers are also 32 bits in size (4 bytes). Pointers are defined by prepending an asterisk (*) to the variable name. Instead of defining a variable of that type, a pointer is defined as something that points to data of that type. The pointer.c program is an example of a pointer being used with the char data type, which is only 1byte in size".
#include <stdio.h>
#include <string.h>
int main()
{
char str_a[20]; // A 20-element character array
char *pointer; // A pointer, meant for a character array
char *pointer2; // And yet another one
strcpy(str_a, "Hello, world!\n");
pointer = str_a; // Set the first pointer to the start of the array.
printf(pointer);
pointer2 = pointer + 2; // Set the second one 2 bytes further in.
printf(pointer2); // Print it.
strcpy(pointer2, "y you guys!\n"); // Copy into that spot.
printf(pointer); // Print again.
}
"As the comments in the code indicate, the first pointer is set at the beginning of the character array. When the character array is referenced like this, it is actually a pointer itself. This is how this buffer was passed as a pointer to the printf() and strcpy() functions earlier. The second pointer is set to the first pointers address plus two, and then some things are printed (shown in the output below)".
reader#hacking:~/booksrc $ gcc -o pointer pointer.c
reader#hacking:~/booksrc $ ./pointer
Hello, world!
llo, world!
Hey you guys!
reader#hacking:~/booksrc $
"The address-of operator is often used in conjunction with pointers, since pointers contain memory addresses. The addressof.c program demonstrates
the address-of operator being used to put the address of an integer variable
into a pointer. This line is shown in bold below".
#include <stdio.h>
int main()
{
int int_var = 5;
int *int_ptr;
int_ptr = &int_var; // put the address of int_var into int_ptr
}
"An additional unary operator called the dereference operator exists for use with pointers. This operator will return the data found in the address the pointer is pointing to, instead of the address itself. It takes the form of an asterisk in front of the variable name, similar to the declaration of a pointer. Once again, the dereference operator exists both in GDB and in C".
"A few additions to the addressof.c code (shown in addressof2.c) will
demonstrate all of these concepts. The added printf() functions use format
parameters, which I’ll explain in the next section. For now, just focus on the programs output".
#include <stdio.h>
int main()
{
int int_var = 5;
int *int_ptr;
int_ptr = &int_var; // Put the address of int_var into int_ptr.
printf("int_ptr = 0x%08x\n", int_ptr);
printf("&int_ptr = 0x%08x\n", &int_ptr);
printf("*int_ptr = 0x%08x\n\n", *int_ptr);
printf("int_var is located at 0x%08x and contains %d\n", &int_var, int_var);
printf("int_ptr is located at 0x%08x, contains 0x%08x, and points to %d\n\n", &int_ptr, int_ptr, *int_ptr);
}
"When the unary operators are used with pointers, the address-of operator can be thought of as moving backward, while the dereference operator moves forward in the direction the pointer is pointing".
Find out more about Pointers & memory allocation
Professor Dan Hirschberg, Computer Science Department, University of California on computer memory https://www.ics.uci.edu/~dan/class/165/notes/memory.html
http://cslibrary.stanford.edu/106/
http://www.programiz.com/c-programming/c-dynamic-memory-allocation
Arrays
Theres a simple tutorial on multi-dimensional arrays by a chap named Alex Allain available here http://www.cprogramming.com/tutorial/c/lesson8.html
Theres information on arrays by a chap named Todd A Gibson available here http://www.augustcouncil.com/~tgibson/tutorial/arr.html
Iterate an Array
#include <stdio.h>
int main()
{
int i;
char char_array[5] = {'a', 'b', 'c', 'd', 'e'};
int int_array[5] = {1, 2, 3, 4, 5};
char *char_pointer;
int *int_pointer;
char_pointer = char_array;
int_pointer = int_array;
for(i=0; i < 5; i++) { // Iterate through the int array with the int_pointer.
printf("[integer pointer] points to %p, which contains the integer %d\n", int_pointer, *int_pointer);
int_pointer = int_pointer + 1;
}
for(i=0; i < 5; i++) { // Iterate through the char array with the char_pointer.
printf("[char pointer] points to %p, which contains the char '%c'\n", char_pointer, *char_pointer);
char_pointer = char_pointer + 1;
}
}
Linked Lists vs Arrays
Arrays are not the only option available, information on Linked List.
http://www.eternallyconfuzzled.com/tuts/datastructures/jsw_tut_linklist.aspx
Conclusion
This information was written simply to pass on some of what I have read throughout my research on the topic that might help others.
Iirc, a string is actually an array of chars, so this should work:
a[1][2]
Quote from the wikipedia article on C pointers -
In C, array indexing is formally defined in terms of pointer arithmetic; that is,
the language specification requires that array[i] be equivalent to *(array + i). Thus in C, arrays can be thought of as pointers to consecutive areas of memory (with no gaps),
and the syntax for accessing arrays is identical for that which can be used to dereference
pointers. For example, an array can be declared and used in the following manner:
int array[5]; /* Declares 5 contiguous (per Plauger Standard C 1992) integers */
int *ptr = array; /* Arrays can be used as pointers */
ptr[0] = 1; /* Pointers can be indexed with array syntax */
*(array + 1) = 2; /* Arrays can be dereferenced with pointer syntax */
So, in response to your question - yes, pointers to pointers can be used as an array without any kind of other declaration at all!
Try a[1][2]. Or *(*(a+1)+2).
Basically, array references are syntactic sugar for pointer dereferencing. a[2] is the same as a+2, and also the same as 2[a] (if you really want unreadable code). An array of strings is the same as a double pointer. So you can extract the second string using either a[1] or *(a+1). You can then find the third character in that string (call it 'b' for now) with either b[2] or *(b + 2). Substituting the original second string for 'b', we end up with either a[1][2] or *(*(a+1)+2).